prove the sequence is increasing
Im asked to show the sequence $a_{n+1}=\sqrt{3+2a_{n}}$ where $a_{1}=0, a_{2}=1$ is increasing and bounded and therefore convergent.
I don't even know how to start the proof. Im sure it increases because its essentially adding a really small number each time and the number gets very small since its being square rooted a bunch of times but i'm having a real issue showing that it actually is increasing for all n.
$$ a_{2}=\sqrt{3} \\ a_{3}=\sqrt{3+2\sqrt{3}} \\ a_{4}=\sqrt{3+2\sqrt{3+2\sqrt{3}}} $$
Everything I try to do leads nowhere and makes no sense. I've seen some proofs that go about by squaring both sides then solving the resulting quadratic equation assuming the sequence has a limit but im not sure how that shows that its increasing. If anyone has any tips that might push my in the right direction it would be greatly appreciated.
$\endgroup$1 Answer
$\begingroup$Note that $a_n \ge 0$ and
$$a_{n+1}^2 - a_n^2 = 3 + 2a_n - 3 -2a_{n-1} = 2(a_n - a_{n-1})$$
thus one can argue by induction that $a_n$ is increasing.
Although you didn't ask, you can also prove by induction that $a_n \le 3$.
$\endgroup$ 5
